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Equal volumes of following solutions are mixed, in which case the pH of resulting solution will be average value of pH of two solutions.
- a)pH = 2 HCl and pH = 12 NaOH
- b)pH = 2 HCl and pH = 4 HCl
- c)pH = 2 HCN and pH = 12 NaOH (Ka of HCN = 10–10)
- d)pH = 5 CH3COOH and pH = 9 NH3(aq), Ka (CH3COOH) = Kb (NH3(aq))
Correct answer is option 'A,D'. Can you explain this answer?
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Equal volumes of following solutions are mixed, in which case the pH o...
(D) pH = 5 CH3COOH and pH = 9 NH3 (aq), both must be of equal concentrations as pKa = pKb. Hence pH = pOH only if concentrations are equal so on mixing equal volume we will get CH3COONH4 salt solution and its pH is given is given by
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Equal volumes of following solutions are mixed, in which case the pH o...
A) pH = 2 HCl and pH = 12 NaOH:
HCl is a strong acid and completely dissociates in water to form H+ and Cl- ions. NaOH is a strong base and completely dissociates in water to form Na+ and OH- ions. When equal volumes of these solutions are mixed, the H+ and OH- ions react to form water, leaving behind Na+ and Cl- ions in the solution. Since NaCl is a salt and does not undergo further dissociation, the resulting solution will be neutral with a pH of 7.
b) pH = 2 HCl and pH = 4 HCl:
Both solutions contain HCl, which is a strong acid. When equal volumes of these solutions are mixed, the resulting solution will still contain HCl and have a pH lower than 4.
c) pH = 2 HCN and pH = 12 NaOH (Ka of HCN = 10^-9.2):
HCN is a weak acid and does not completely dissociate in water. NaOH is a strong base and completely dissociates in water to form Na+ and OH- ions. When equal volumes of these solutions are mixed, the OH- ions react with the HCN molecules to form CN- ions and water. The resulting solution will be basic since the concentration of OH- ions is higher than the concentration of H+ ions. To calculate the pH of the resulting solution, we need to find the concentration of CN- ions.
HCN + OH- → CN- + H2O
At equilibrium, the following expression holds true:
Ka = [H+][CN-]/[HCN]
Since the solution is at equilibrium, the concentration of CN- ions is equal to the concentration of HCN molecules that have reacted with OH- ions. Let x be the concentration of CN- ions in the resulting solution. Then, the concentration of HCN molecules that have reacted with OH- ions is also x.
Ka = [H+][CN-]/[HCN]
10^-9.2 = x^2/(0.5-x)
Assuming x is much smaller than 0.5, we can simplify the expression:
10^-9.2 = x^2/0.5
x = 1.4 x 10^-5 M
The concentration of OH- ions in the resulting solution is also 1.4 x 10^-5 M. Therefore, the pOH of the resulting solution is:
pOH = -log(1.4 x 10^-5) = 4.85
The pH of the resulting solution can be calculated using the following equation:
pH + pOH = 14
pH = 14 - 4.85 = 9.15
Therefore, the pH of the resulting solution will be 9.15, which is the average value of the pH of the two solutions.
HCl is a strong acid and completely dissociates in water to form H+ and Cl- ions. NaOH is a strong base and completely dissociates in water to form Na+ and OH- ions. When equal volumes of these solutions are mixed, the H+ and OH- ions react to form water, leaving behind Na+ and Cl- ions in the solution. Since NaCl is a salt and does not undergo further dissociation, the resulting solution will be neutral with a pH of 7.
b) pH = 2 HCl and pH = 4 HCl:
Both solutions contain HCl, which is a strong acid. When equal volumes of these solutions are mixed, the resulting solution will still contain HCl and have a pH lower than 4.
c) pH = 2 HCN and pH = 12 NaOH (Ka of HCN = 10^-9.2):
HCN is a weak acid and does not completely dissociate in water. NaOH is a strong base and completely dissociates in water to form Na+ and OH- ions. When equal volumes of these solutions are mixed, the OH- ions react with the HCN molecules to form CN- ions and water. The resulting solution will be basic since the concentration of OH- ions is higher than the concentration of H+ ions. To calculate the pH of the resulting solution, we need to find the concentration of CN- ions.
HCN + OH- → CN- + H2O
At equilibrium, the following expression holds true:
Ka = [H+][CN-]/[HCN]
Since the solution is at equilibrium, the concentration of CN- ions is equal to the concentration of HCN molecules that have reacted with OH- ions. Let x be the concentration of CN- ions in the resulting solution. Then, the concentration of HCN molecules that have reacted with OH- ions is also x.
Ka = [H+][CN-]/[HCN]
10^-9.2 = x^2/(0.5-x)
Assuming x is much smaller than 0.5, we can simplify the expression:
10^-9.2 = x^2/0.5
x = 1.4 x 10^-5 M
The concentration of OH- ions in the resulting solution is also 1.4 x 10^-5 M. Therefore, the pOH of the resulting solution is:
pOH = -log(1.4 x 10^-5) = 4.85
The pH of the resulting solution can be calculated using the following equation:
pH + pOH = 14
pH = 14 - 4.85 = 9.15
Therefore, the pH of the resulting solution will be 9.15, which is the average value of the pH of the two solutions.
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Equal volumes of following solutions are mixed, in which case the pH of resulting solution will be average value of pH of two solutions.a)pH = 2 HCl and pH = 12 NaOHb)pH = 2 HCl and pH = 4 HClc)pH = 2 HCN and pH = 12 NaOH (Ka of HCN = 10–10) d)pH = 5 CH3COOH and pH = 9 NH3(aq), Ka (CH3COOH) = Kb (NH3(aq))Correct answer is option 'A,D'. Can you explain this answer?
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Equal volumes of following solutions are mixed, in which case the pH of resulting solution will be average value of pH of two solutions.a)pH = 2 HCl and pH = 12 NaOHb)pH = 2 HCl and pH = 4 HClc)pH = 2 HCN and pH = 12 NaOH (Ka of HCN = 10–10) d)pH = 5 CH3COOH and pH = 9 NH3(aq), Ka (CH3COOH) = Kb (NH3(aq))Correct answer is option 'A,D'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Equal volumes of following solutions are mixed, in which case the pH of resulting solution will be average value of pH of two solutions.a)pH = 2 HCl and pH = 12 NaOHb)pH = 2 HCl and pH = 4 HClc)pH = 2 HCN and pH = 12 NaOH (Ka of HCN = 10–10) d)pH = 5 CH3COOH and pH = 9 NH3(aq), Ka (CH3COOH) = Kb (NH3(aq))Correct answer is option 'A,D'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Equal volumes of following solutions are mixed, in which case the pH of resulting solution will be average value of pH of two solutions.a)pH = 2 HCl and pH = 12 NaOHb)pH = 2 HCl and pH = 4 HClc)pH = 2 HCN and pH = 12 NaOH (Ka of HCN = 10–10) d)pH = 5 CH3COOH and pH = 9 NH3(aq), Ka (CH3COOH) = Kb (NH3(aq))Correct answer is option 'A,D'. Can you explain this answer?.
Equal volumes of following solutions are mixed, in which case the pH of resulting solution will be average value of pH of two solutions.a)pH = 2 HCl and pH = 12 NaOHb)pH = 2 HCl and pH = 4 HClc)pH = 2 HCN and pH = 12 NaOH (Ka of HCN = 10–10) d)pH = 5 CH3COOH and pH = 9 NH3(aq), Ka (CH3COOH) = Kb (NH3(aq))Correct answer is option 'A,D'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Equal volumes of following solutions are mixed, in which case the pH of resulting solution will be average value of pH of two solutions.a)pH = 2 HCl and pH = 12 NaOHb)pH = 2 HCl and pH = 4 HClc)pH = 2 HCN and pH = 12 NaOH (Ka of HCN = 10–10) d)pH = 5 CH3COOH and pH = 9 NH3(aq), Ka (CH3COOH) = Kb (NH3(aq))Correct answer is option 'A,D'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Equal volumes of following solutions are mixed, in which case the pH of resulting solution will be average value of pH of two solutions.a)pH = 2 HCl and pH = 12 NaOHb)pH = 2 HCl and pH = 4 HClc)pH = 2 HCN and pH = 12 NaOH (Ka of HCN = 10–10) d)pH = 5 CH3COOH and pH = 9 NH3(aq), Ka (CH3COOH) = Kb (NH3(aq))Correct answer is option 'A,D'. Can you explain this answer?.
Solutions for Equal volumes of following solutions are mixed, in which case the pH of resulting solution will be average value of pH of two solutions.a)pH = 2 HCl and pH = 12 NaOHb)pH = 2 HCl and pH = 4 HClc)pH = 2 HCN and pH = 12 NaOH (Ka of HCN = 10–10) d)pH = 5 CH3COOH and pH = 9 NH3(aq), Ka (CH3COOH) = Kb (NH3(aq))Correct answer is option 'A,D'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
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Equal volumes of following solutions are mixed, in which case the pH of resulting solution will be average value of pH of two solutions.a)pH = 2 HCl and pH = 12 NaOHb)pH = 2 HCl and pH = 4 HClc)pH = 2 HCN and pH = 12 NaOH (Ka of HCN = 10–10) d)pH = 5 CH3COOH and pH = 9 NH3(aq), Ka (CH3COOH) = Kb (NH3(aq))Correct answer is option 'A,D'. Can you explain this answer?, a detailed solution for Equal volumes of following solutions are mixed, in which case the pH of resulting solution will be average value of pH of two solutions.a)pH = 2 HCl and pH = 12 NaOHb)pH = 2 HCl and pH = 4 HClc)pH = 2 HCN and pH = 12 NaOH (Ka of HCN = 10–10) d)pH = 5 CH3COOH and pH = 9 NH3(aq), Ka (CH3COOH) = Kb (NH3(aq))Correct answer is option 'A,D'. Can you explain this answer? has been provided alongside types of Equal volumes of following solutions are mixed, in which case the pH of resulting solution will be average value of pH of two solutions.a)pH = 2 HCl and pH = 12 NaOHb)pH = 2 HCl and pH = 4 HClc)pH = 2 HCN and pH = 12 NaOH (Ka of HCN = 10–10) d)pH = 5 CH3COOH and pH = 9 NH3(aq), Ka (CH3COOH) = Kb (NH3(aq))Correct answer is option 'A,D'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice Equal volumes of following solutions are mixed, in which case the pH of resulting solution will be average value of pH of two solutions.a)pH = 2 HCl and pH = 12 NaOHb)pH = 2 HCl and pH = 4 HClc)pH = 2 HCN and pH = 12 NaOH (Ka of HCN = 10–10) d)pH = 5 CH3COOH and pH = 9 NH3(aq), Ka (CH3COOH) = Kb (NH3(aq))Correct answer is option 'A,D'. Can you explain this answer? tests, examples and also practice JEE tests.
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